Physics, 20.07.2021 03:20, andrewmena05

A projectile is shot from the edge of a cliff 125 m above ground level with an initial speed of 65.0 m/s at an angle of 37 degrees with the horizontal. (a) determine the time taken by the projectile to hit the point P at ground level. (b) determine the range X of the projectile as measured from the base of the cliff at the instant just before the projectile hits point P. Find (x) the horizontal and vertical components of its velocity and (d) the magnitude of the velocity and (e) the angle made by the velocity vector with the horizontal (f) Find the maximum height above the cliff top reached by the projectile

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answered: Emptypockets451

(a) The time it takes the projectile to hit the ground is approximately 10.42 seconds

(b) The range of the projectile from the base of the cliff is approximately 540.92 meters

The vertical component of the velocity is approximately 39.12 m/s

(d) The magnitude of the velocity of the projectile is 65.0 m/s

(e) The angle made by the velocity vector of the projectile with the horizontal is 37°

(f) The maximum height above the cliff top reached by the projectile is approximately 77.99 meters

Explanation:

The height of the cliff above ground level, y₀ = 125 m

The initial speed of the projectile, u = 65.0 m/s

The angle of elevation of the projectile, θ = 37°

(a) The time for the projectile to hit the ground, t, is given as follows;

y = y₀ + u·sin(θ)·t -(1/2)·g·t²

g = 9.81 m/s²

At ground level, y = 0, we get;

0 = 125 + 65·sin(37)·t - 4.905·t²

t = (-65·sin(37) ± √((65·sin(37))² - 4×(-4.905)×125))/(2 × (-4.905))

∴ t ≈ -2.45 or t ≈ 10.42

The time it takes the projectile to hit the ground, t ≈ 10.42 seconds

(b) The range of the projectile X = u·cos(θ) × t

∴ X = 65.0 × cos(37°) × 10.42 ≈ 540.92

The range of the projectile from the base of the cliff, X ≈ 540.92 meters

∴ uₓ = 65.0 ×cos(37°) ≈ 51.91

The horizontal component of the velocity, uₓ ≈ 51.91 m/s

The vertical component of the velocity, u_y = u×sin(θ)

∴ u_y = 65.0 ×sin(37°) ≈ 39.12

The vertical component of the velocity, u_y ≈ 39.12 m/s

(d) The magnitude of the velocity, $\left | u \right |$ , is the given speed = 65.0 m/s

$\left | u \right |$ = √( u_y ² + uₓ²)

∴ $\left | u \right |$ = √(39.12² + 51.91²) = 65

$\left | u \right |$ = 65.0 m/s

(e) The angle made by the velocity vector with the horizontal = The angle of elevation with which the projectile is launched = 37°

(f) The maximum height above the cliff top reached by the projectile, $y_{max \ of \ cliff}$ , is given as follows;

$y_{max \ of \ cliff}= \dfrac{ \left(u \times sin(\theta)\right)^2}{2 \cdot g}$

∴ $y_{max \ of \ cliff}$ = (65.0 × sin(37°))²/(2 × 9.81) ≈ 77.99

$y_{max \ of \ cliff}$ ≈ 77.99 meters.

answered: Guest

the independent variable, the one that is intentionally manipulated, is whether or not the car hits the barrier.

the dependent variable, the one that you measure the response in, is the distance that the washers travel.

answered: Guest

russian is not my first language, so i'm afraid i can't write out an answer in russian, but hopefully you can find a way to do so. here's my best translation of the problem: you're given a plot of a particle's position in one-dimensional space over time, and you're asked to find either the total distance traveled by the particle after 6 seconds had elapsed, or the total displacement of the particle after 6 seconds.

if you want the total distance traveled, you would first need to determine the particle's velocity as a function of time, then integrate that over the 6-second interval. we know the position function is quadratic with roots at 0 and 6, so we have

$x(t)=ax(x-6)$

where $a$ is unknown. because the position is parabolic, there is symmetry about the midpoint at $t=3$ , and we know that

$x(3)=10\implies -9a=10\implies a=-\dfrac{10}9$

so that the position function is

$x(t)=-\dfrac{10}9x(x-6)$

which means the velocity is

$v(t)=\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac{20}9x+\dfrac{60}9$

then the total distance traveled is

$\displaystyle\int_{t=0}^{t=6}|v(t)|\,\mathrm dt=\left\{\int_{t-0}^{t=3}-\int_{t=3}^{t=6}\right\}v(t)\,\mathrm dt=20$

so the total distance traveled is 20 m.

on the other hand, if you want to find displacement: the particle starts at $x=0$ and ends at $x=0$ , so its displacement is 0.

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A projectile is shot from the edge of a cliff 125 m above ground level with an initial speed of 65.0...